I'm setting up a script and I want it to systematically go through ALL POSSIBLE 2x2, 3x3, and 4x4 matrices modulo 2, 3, 4, 5, 6, and 7. For example, for modulus 2 in a 2x2, there would be 16 possibilities (4^2 because there are 4 positions with 2 possibilities each). I'm having trouble getting MATLAB to not only form all these possibilities but to put them through my script one at a time. Any thoughts?
Thanks!
This solution uses allcomb from matlab file exchange.
%size
n=2
%maximum value
m=2
%generate input for allcomb
e=cell(1,n^2)
e(1:end)={[0:m-1]}
%generate all combinations.
F=reshape(allcomb(e{:}),[],n,n)
F is a 3D-Matrix, to get the first possibility use:
squeeze(F(1,:,:))
A slight generalization of this Q&A does the job in one line:
r = 2; %// number of rows
c = 2; %// number of columns
n = 2; %// considered values: 0, 1, ..., n-1
M = reshape(dec2base(0:n^(r*c)-1, n).' - '0', r,c,[]);
Result for r, c, n as above:
M(:,:,1) =
0 0
0 0
M(:,:,2) =
0 0
0 1
...
M(:,:,16) =
1 1
1 1
Related
I have a matrix index in Matlab with size GxN and a matrix A with size MxN.
Let me provide an example before presenting my question.
clear
N=3;
G=2;
M=5;
index=[1 2 3;
13 14 15]; %GxN
A=[1 2 3;
5 6 7;
21 22 23;
1 2 3;
13 14 15]; %MxN
I would like your help to construct a matrix Response with size GxM with Response(g,m)=1 if the row A(m,:) is equal to index(g,:) and zero otherwise.
Continuing the example above
Response= [1 0 0 1 0;
0 0 0 0 1]; %GxM
This code does what I want (taken from a previous question of mine - just to clarify: the current question is different)
Response=permute(any(all(bsxfun(#eq, reshape(index.', N, [], G), permute(A, [2 3 4 1])), 1), 2), [3 4 1 2]);
However, the command is extremely slow for my real matrix sizes (N=19, M=500, G=524288). I understand that I will not be able to get huge speed but anything that can improve on this is welcome.
Aproach 1: computing distances
If you have the Statistics Toolbox:
Response = ~(pdist2(index, A));
or:
Response = ~(pdist2(index, A, 'hamming'));
This works because pdist2 computes the distance between each pair of rows. Equal rows have distance 0. The logical negation ~ gives 1 for those pairs of rows, and 0 otherwise.
Approach 2: reducing rows to unique integer labels
This approach is faster on my machine:
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
It works by reducing rows to unique integer labels (using the third output of unique), and comparing the latter instead of the former.
For your size values this takes approximately 1 second on my computer:
clear
N = 19; M = 500; G = 524288;
index = randi(5,G,N); A = randi(5,M,N);
tic
[~,~,u] = unique([index; A], 'rows');
Response = bsxfun(#eq, u(1:G), u(G+1:end).');
toc
gives
Elapsed time is 1.081043 seconds.
MATLAB has a multitude of functions for working with sets, including setdiff, intersect, union etc. In this case, you can use the ismember function:
[~, Loc] = ismember(A,index,'rows');
Which gives:
Loc =
1
0
0
1
2
And Response would be constructed as follows:
Response = (1:size(index,1) == Loc).';
Response =
2×5 logical array
1 0 0 1 0
0 0 0 0 1
You could reshape the matrices so that each row instead lies along the 3rd dimension. Then we can use implicit expansion (see bsxfun for R2016b or earlier) for equality of all elements, and all to aggregate on the rows (i.e. false if not all equal for a given row).
Response = all( reshape( index, [], 1, size(index,2) ) == reshape( A, 1, [], size(A,2) ), 3 );
You might even be able to avoid some reshaping by using all in another dimension, but it's easier for me to visualise it this way.
I have a matrix suppX in Matlab with size GxN and a matrix A with size MxN. I would like your help to construct a matrix Xresponse with size GxM with Xresponse(g,m)=1 if the row A(m,:) is equal to the row suppX(g,:) and zero otherwise.
Let me explain better with an example.
suppX=[1 2 3 4;
5 6 7 8;
9 10 11 12]; %GxN
A=[1 2 3 4;
1 2 3 4;
9 10 11 12;
1 2 3 4]; %MxN
Xresponse=[1 1 0 1;
0 0 0 0;
0 0 1 0]; %GxM
I have written a code that does what I want.
Xresponsemy=zeros(size(suppX,1), size(A,1));
for x=1:size(suppX,1)
Xresponsemy(x,:)=ismember(A, suppX(x,:), 'rows').';
end
My code uses a loop. I would like to avoid this because in my real case this piece of code is part of another big loop. Do you have suggestions without looping?
One way to do this would be to treat each matrix as vectors in N dimensional space and you can find the L2 norm (or the Euclidean distance) of each vector. After, check if the distance is 0. If it is, then you have a match. Specifically, you can create a matrix such that element (i,j) in this matrix calculates the distance between row i in one matrix to row j in the other matrix.
You can treat your problem by modifying the distance matrix that results from this problem such that 1 means the two vectors completely similar and 0 otherwise.
This post should be of interest: Efficiently compute pairwise squared Euclidean distance in Matlab.
I would specifically look at the answer by Shai Bagon that uses matrix multiplication and broadcasting. You would then modify it so that you find distances that would be equal to 0:
nA = sum(A.^2, 2); % norm of A's elements
nB = sum(suppX.^2, 2); % norm of B's elements
Xresponse = bsxfun(#plus, nB, nA.') - 2 * suppX * A.';
Xresponse = Xresponse == 0;
We get:
Xresponse =
3×4 logical array
1 1 0 1
0 0 0 0
0 0 1 0
Note on floating-point efficiency
Because you are using ismember in your implementation, it's implicit to me that you expect all values to be integer. In this case, you can very much compare directly with the zero distance without loss of accuracy. If you intend to move to floating-point, you should always compare with some small threshold instead of 0, like Xresponse = Xresponse <= 1e-10; or something to that effect. I don't believe that is needed for your scenario.
Here's an alternative to #rayryeng's answer: reduce each row of the two matrices to a unique identifier using the third output of unique with the 'rows' input flag, and then compare the identifiers with singleton expansion (broadcast) using bsxfun:
[~, ~, w] = unique([A; suppX], 'rows');
Xresponse = bsxfun(#eq, w(1:size(A,1)).', w(size(A,1)+1:end));
New to MatLab here (R2015a, Mac OS 10.10.5), and hoping to find a solution to this indexing problem.
I want to change the values of a large 2D matrix, based on one vector of row indices and one of column indices. For a very simple example, if I have a 3 x 2 matrix of zeros:
A = zeros(3, 2)
0 0
0 0
0 0
I want to change A(1, 1) = 1, and A(2, 2) = 1, and A(3, 1) = 1, such that A is now
1 0
0 1
1 0
And I want to do this using vectors to indicate the row and column indices:
rows = [1 2 3];
cols = [1 2 1];
Is there a way to do this without looping? Remember, this is a toy example that needs to work on a very large 2D matrix. For extra credit, can I also include a vector that indicates which value to insert, instead of fixing it at 1?
My looping approach is easy, but slow:
for i = 1:length(rows)
A(rows(i), cols(i)) = 1;
end
sub2ind can help here,
A = zeros(3,2)
rows = [1 2 3];
cols = [1 2 1];
A(sub2ind(size(A),rows,cols))=1
A =
1 0
0 1
1 0
with a vector to 'insert'
b = [1,2,3];
A(sub2ind(size(A),rows,cols))=b
A =
1 0
0 2
3 0
I found this answer online when checking on the speed of sub2ind.
idx = rows + (cols - 1) * size(A, 1);
therefore
A(idx) = 1 % or b
5 tests on a big matrix (~ 5 second operations) shows it's 20% faster than sub2ind.
There is code for an n-dimensional problem here too.
What you have is basically a sparse definition of a matrix. Thus, an alternative to sub2ind is sparse. It will create a sparse matrix, use full to convert it to a full matrix.
A=full(sparse(rows,cols,1,3,2))
I am confused by the
[m,n]=hist(y,x)
such as
M = [1, 2, 3;
4, 5, 6;
1, 2, 3];
[m,n] = hist(M,1:3)
Which results in
m = 2 0 0
0 2 0
1 1 3
Can someone please explain how m is calculated?
hist actually takes vectors as input arguments, you wrote a matrix, so it just handles your input as if it was several vector-inputs. The output are the number of elements for each container (in your case 1:3, the second argument).
[m,n] = hist([1,2,3;4,5,6;1,2,3],1:3)
treats each column as one input. You put in 3 inputs (# of columns) and you get 3 outputs.
[2 0 1]'
means, for the input [1;4;1] and the bin 1:3 two elements are in bin 1 and one element is in bin 3.
Look at the last column of m, here all three values are in the third bin, which makes sense, since the corresponding vector is [3;6;3], and out of those numbers all have to go into the bin/container 3.
Hi I have the following matrix:
A= 1 2 3;
0 4 0;
1 0 9
I want matrix A to be:
A= 1 2 3;
1 4 9
PS - semicolon represents the end of each column and new column starts.
How can I do that in Matlab 2014a? Any help?
Thanks
The problem you run into with your problem statement is the fact that you don't know the shape of the "squeezed" matrix ahead of time - and in particular, you cannot know whether the number of nonzero elements is a multiple of either the rows or columns of the original matrix.
As was pointed out, there is a simple function, nonzeros, that returns the nonzero elements of the input, ordered by columns. In your case,
A = [1 2 3;
0 4 0;
1 0 9];
B = nonzeros(A)
produces
1
1
2
4
3
9
What you wanted was
1 2 3
1 4 9
which happens to be what you get when you "squeeze out" the zeros by column. This would be obtained (when the number of zeros in each column is the same) with
reshape(B, 2, 3);
I think it would be better to assume that the number of elements may not be the same in each column - then you need to create a sparse array. That is actually very easy:
S = sparse(A);
The resulting object S is a sparse array - that is, it contains only the non-zero elements. It is very efficient (both for storage and computation) when lots of elements are zero: once more than 1/3 of the elements are nonzero it quickly becomes slower / bigger. But it has the advantage of maintaining the shape of your matrix regardless of the distribution of zeros.
A more robust solution would have to check the number of nonzero elements in each column and decide what the shape of the final matrix will be:
cc = sum(A~=0);
will count the number of nonzero elements in each column of the matrix.
nmin = min(cc);
nmax = max(cc);
finds the smallest and largest number of nonzero elements in any column
[i j s] = find(A); % the i, j coordinates and value of nonzero elements of A
nc = size(A, 2); % number of columns
B = zeros(nmax, nc);
for k = 1:nc
B(1:cc(k), k) = s(j == k);
end
Now B has all the nonzero elements: for columns with fewer nonzero elements, there will be zero padding at the end. Finally you can decide if / how much you want to trim your matrix B - if you want to have no zeros at all, you will need to trim some values from the longer columns. For example:
B = B(1:nmin, :);
Simple solution:
A = [1 2 3;0 4 0;1 0 9]
A =
1 2 3
0 4 0
1 0 9
A(A==0) = [];
A =
1 1 2 4 3 9
reshape(A,2,3)
ans =
1 2 3
1 4 9
It's very simple though and might be slow. Do you need to perform this operation on very large/many matrices?
From your question it's not clear what you want (how to arrange the non-zero values, specially if the number of zeros in each column is not the same). Maybe this:
A = reshape(nonzeros(A),[],size(A,2));
Matlab's logical indexing is extremely powerful. The best way to do this is create a logical array:
>> lZeros = A==0
then use this logical array to index into A and delete these zeros
>> A(lZeros) = []
Finally, reshape the array to your desired size using the built in reshape command
>> A = reshape(A, 2, 3)